Solve the wave equation $ (\partial^2_t- \nabla^2) u(t,\mathbf{x}) = f(t,\mathbf{x})$ for a source that is always there Consider the inhomogeneous wave-equation (with units $c=1$ and in 3 spatial dimensions)
$$
 \frac{\partial^2 u}{\partial t^2} - \nabla^2 u(t,\mathbf{x}) = f(t,\mathbf{x}) \tag{1}
$$
where $f(t,\mathbf{x})$ is a source, which is non-zero for all $t>0$. How would one find a solution to this equation if the source is always there?
More Commonly: What I have often seen in the literature is a solution to the above equation using the retarded propagator
$$
G_R(t,\mathbf{x}; t', \mathbf{x}') = \frac{\Theta(t - t')\delta(t - t' - |\mathbf{x} - \mathbf{x}'|)}{4\pi |\mathbf{x} - \mathbf{x}'|}
$$
which is usually used for a similar problem where the source ``turns on'' at some time, say $t=0$, so that the wave equation is instead $(\partial_{t}^2 - \nabla^2 ) u(t,\mathbf{x}) = \Theta(t) f(t,\mathbf{x})$. Usually $u(t,\mathbf{x})$ is given a set of boundary conditions for $t<0$, such that $u(t,\mathbf{x}) = u_0(t,\mathbf{x})$ for $t<0$ (and a time-derivative condition like $\partial_t u(t,\mathbf{x}) = \partial_t u_0(t,\mathbf{x})$ for $t<0$). Then the solution for the source that turns on has the form
$$
u(t,\mathbf{x}) = u_0(t,\mathbf{x}) + \int_0^\infty dt' \int d^{3}\mathbf{x}' \; G_{R}(t,\mathbf{x} ; t', \mathbf{x}') f(t',\mathbf{x}')
$$
My Question: How do you solve (1) given that the source $f$ is there for all $-\infty < t < \infty$? This would probably require giving $u(t,\mathbf{x})$ some sort of different boundary conditions, like $u(t,\mathbf{x}) |_{\mathbf{x} \in S} = s(t,\mathbf{x})$ along some surface $S$ (but this is just a guess).
EDIT: For example one could consider a point charge that sits at the origin forever with $f(t,\mathbf{x}) = - \nabla \rho(\mathbf{x})$ where the charge density is $\rho = q \delta(\mathbf{x})$ for some constant $q>0$.
 A: Typically the solution for the inhomogeneous problem is obtained via the homogeneous problem via Duhamel's principle. In short, the solution to your equation (with zero initial conditions) is
$$
u(t, x) = \int_0^t v(t, x; s)\, ds
$$
where $v(\cdot, \cdot; s) : (0, \infty) \times \mathbb{R}^3 \to \mathbb{R}$ is the solution to the equation
$$
\begin{cases}
(\partial_t^2 - \Delta)v(t, x; s) = 0 \\
v(t, x; s) = 0, v_t(t, x;s) = f(x, s). \tag{$1s$}
\end{cases}
$$
Note that in 3D this $v$ is given by
$$
v(t, x; s) = \frac{1}{|\partial B(x, t)|}\int_{\partial B(x,t)} th(y)\, dS(y).
$$
Therefore
$$
u(t, x) = \int_0^t \frac{1}{|\partial B(x, t)|}\int_{\partial B(x,t)} th(y)\, dS(y)\, dt.
$$
See Evans for more details. One can formulate this question with wave propagators as well; in that case the solution is
$$
u(t, x) = \cos(t\sqrt{-\Delta})u_0(0,x) + \frac{\sin(t\sqrt{-\Delta})}{\sqrt{-\Delta}}u_1(0, x) - \int_0^t \frac{\sin((t - s)\sqrt{-\Delta})}{\sqrt{-\Delta}}F(s)\, ds.
$$
(See Exercise 2.22 of Tao's book on nonlinear dispersive equations.) Here $u_0(x) = u(0, x)$ and $u_1(x) = \partial_t u(0, x)$ are general initial conditions.
